The system shown is at rest when a constant 250N force is applied to block A. Neglecting masses of pulleys and friction, determine:
a) Velocity of block B after block A moved 2 m.
b) The tension in the cable
Fnet = ma.
U = mgh
KE = 1/2 mv^2
Ui + KEi + W = Uf + KEf, where subindex i and f mean initial and final states.
The Attempt at a Solution
I have completed this problem, but I'm not sure if this is correct. The professor said this particular problem can be done by taking both masses and the ropes as a single system and also by taking each mass as its own, separate system. I did took each as a separate system.
First, I drew a free body diagram for block A. The net force in the x direction, assuming left is negative gives me:
F - T = ma_A
1------------------250N - T = ma_A
The relationship between the position of block A and block B was obtained and determined to be:
X_A + 3X_B = constant. Taking first time derivative:
V_A + 3V_B = 0
V_A = - 3V_B. Second time derivative:
2--------------a_A = -3a_B
Using newton's law for the y direction on block B:
up positive: 3T - W = m*a_B
3-----------3T - mg = m*a_B
Solving 1 for acceleration a_A, then using equation 2 to leave as a_B = form:
250-T/m_a = -3a_B
4----- -250+T/(3m_a) = a_B
Using 4 to solve for tension in 3:
3T - m_b*g = -m_b*250/(3*m_a) + m_b * T/(3m_a)
Subsituting for values of mass of b, g, mass of a and solving for T:
T(3-25/90) = 25*9.81 - 25*250/(3*30)
T = 64.6N
Then, with that tension, using the work energy theorem on block A:
Wtension + Wconst force = 1/2 m_a V_a^2
250*2 - 129.2 = 1/2 *30va^2
V_a = 4.97 m /s <------ left.
V_a = -3Vb
V_b = 1.66 m / s Upwards.
Does that seem right...? Thanks a ton. :]